Suppose $M$ and $N$ are smooth manifolds and $F:M\to N$ a smooth map. Assume that $dF_p:T_pM \to T_{F(p)}N$ is the zero map for each $p \in M$. Show that $F$ is constant.
Letting $(U,\varphi)$ to be a chart on $M$ and $(V, \psi)$ be a chart on $N$ I have the coorindate representation $\hat{F}:=\psi \circ F \circ \varphi^{-1}:\varphi(U \cap F^{-1}(V)) \to \psi(V)$.
Lee, then suggests that $dF_p( \frac{\partial}{\partial x^i} \bigg|_p)=\frac{\partial \hat{F}}{\partial x^i} (\hat{p})\frac{\partial}{\partial y^j} \bigg|_{F(p)}$, but if the differential is the zero map, then $$dF_p( \frac{\partial}{\partial x^i} \bigg|_p)=\frac{\partial \hat{F}}{\partial x^i} (\hat{p})\frac{\partial}{\partial y^j} \bigg|_{F(p)}=0.$$
How does this imply that $\hat{F}$ would be constant? I'm considering an arbitary basis vector under the differential, but what if I considered something otherthan a basis vector?
Best Answer
You need $M$ to be connected for the result to hold. If two points $x$ and $y$ are close enough, you can work in coordinate charts and you are reduced to the case of a map between ${\bf R}^n$ to ${\bf R}^m$. There it follows from a standard estimate that comes from the Taylor expansion.